Show $\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$

Question

I have some difficulty to prove the following limit: $$\lim_{N\to \infty}\sum_{k=1}^{N}\frac{1}{k+N}=\ln(2)$$ Can someone help me? Thanks.

Answer 1

Hint: It is a Riemann Sum that corresponds to the integral $$\int_{0}^1 \frac{1}{1+x} dx,$$ and this evaluates to $\ln 2$.

Answer 2

A common estimate for the Harmonic Numbers is $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+O\left(\frac1n\right)\tag{1} $$ where $\gamma$ is the Euler-Mascheroni constant.

Applying $(1)$, we get that $$ \begin{align} \sum_{k=1}^{N}\frac{1}{k+N} &=\sum_{k=1}^{2N}\frac1k-\sum_{k=1}^N\frac1k\\ &=\left(\log(2N)+\gamma+O\left(\frac{1}{2N}\right)\right)-\left(\log(N)+\gamma+O\left(\frac1N\right)\right)\\ &=\log(2)+O\left(\frac1N\right)\tag{2} \end{align} $$ Taking the limit of $(2)$ as $N\to\infty$ yields $$ \lim_{N\to\infty}\sum_{k=1}^{N}\frac{1}{k+N}=\log(2)\tag{3} $$

Answer 3

$$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$$

You are looking for $f(1)$

$$f(x)=\lim_{n\to\infty}\frac{x}{n} \sum \limits_{k=1}^n \frac{1}{1+\frac{kx}{n}}=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n (1-\frac{kx}{n}+\frac{k^2x^2}{n^2}-\frac{k^3x^3}{n^3}+....)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....$$

$$f(x)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....=\lim_{n\to\infty} \frac{x}{n} n -\lim_{n\to\infty} \frac{x^2}{n^2} (\frac{n^2}{2}+\frac{n}{2})+\lim_{n\to\infty} \frac{x^3}{n^3} (\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})-\lim_{n\to\infty} \frac{x^4}{n^4}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+....$$

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$$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants.where aj are constants. More information about summation http://en.wikipedia.org/wiki/Summation

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After solving limits. We get:

$$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$$

$$f(x)=\ln(x+1)$$ $$f(1)=\ln(2)$$

Answer 4

I just found this question and thought it might be instructive to present a way forward that relies on the Taylor series for $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k}$ from which we see that $\log(2)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$. To that end, we proceed.

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Note that we can write

$$\begin{align} \sum_{k=1}^N \frac{1}{k+N}&=\sum_{k=N+1}^{2N}\frac1k\\\\ &=\sum_{k=1}^{2N}\frac1k-\sum_{k=1}^N\frac1k \tag 1 \end{align}$$

Next, we split the first sum on the right-hand side into sums of the even terms and odd terms and find that

$$\begin{align} \color{blue}{\sum_{k=1}^{2N}\frac1k}-\color{red}{\sum_{k=1}^N\frac1k}&=\color{blue}{\left(\sum_{k=1}^N \frac{1}{2k}+\sum_{k=1}^N \frac{1}{2k-1}\right)}-\color{red}{\sum_{k=1}^N\frac1k}\\\\ &=\sum_{k=1}^N \frac{1}{2k-1}-\sum_{k=1}^N \frac{1}{2k}\\\\ &=\sum_{k=1}^{2N}\frac{(-1)^{k-1}}{k}\tag2 \end{align}$$

Letting $N\to \infty$ in $(2)$, we find

$$\begin{align} \lim_{N\to \infty}\sum_{k=1}^N \frac{1}{k+N}&=\lim_{N\to \infty}\sum_{k=1}^{2N}\frac{(-1)^{k-1}}{k}\\\\ &=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\\\\ &=\log(2) \end{align}$$

as was to be shown!